Key Concepts and Formulas

• Infinite Geometric Series: The sum of an infinite geometric series with first term $a$ and common ratio $r$ is given by $S = \frac{a}{1-r}$, provided $|r| < 1$.
• Splitting Series: An infinite series can be split into sub-series based on the parity of the index (even or odd terms).
• Properties of $(-1)^n$: The term $(-1)^n$ alternates between $-1$ for odd $n$ and $1$ for even $n$.

Step-by-Step Solution

Step 1: Analyze the terms of the series A and B. We are given two series, $A$ and $B$. The general term in both series depends on $(3 + (-1)^n)^n$. The behavior of $(-1)^n$ changes based on whether $n$ is even or odd. Let's evaluate the denominator for even and odd $n$: If $n$ is odd, $n = 2k-1$ for $k \ge 1$. Then $(-1)^n = -1$, so $3 + (-1)^n = 3 - 1 = 2$. If $n$ is even, $n = 2k$ for $k \ge 1$. Then $(-1)^n = 1$, so $3 + (-1)^n = 3 + 1 = 4$.

Step 2: Expand the series A and B by separating even and odd terms. For series A: The terms where $n$ is odd ($n=1, 3, 5, \dots$) are $\frac{1}{(3+(-1)^n)^n} = \frac{1}{2^n}$. The terms where $n$ is even ($n=2, 4, 6, \dots$) are $\frac{1}{(3+(-1)^n)^n} = \frac{1}{4^n}$. So, $A = \sum_{n=1}^\infty \frac{1}{(3+(-1)^n)^n} = \left(\frac{1}{2^1} + \frac{1}{2^3} + \frac{1}{2^5} + \dots\right) + \left(\frac{1}{4^2} + \frac{1}{4^4} + \frac{1}{4^6} + \dots\right)$.

For series B: The terms where $n$ is odd ($n=1, 3, 5, \dots$) are $\frac{(-1)^n}{(3+(-1)^n)^n} = \frac{-1}{2^n}$. The terms where $n$ is even ($n=2, 4, 6, \dots$) are $\frac{(-1)^n}{(3+(-1)^n)^n} = \frac{1}{4^n}$. So, $B = \sum_{n=1}^\infty \frac{(-1)^n}{(3+(-1)^n)^n} = \left(\frac{-1}{2^1} + \frac{-1}{2^3} + \frac{-1}{2^5} + \dots\right) + \left(\frac{1}{4^2} + \frac{1}{4^4} + \frac{1}{4^6} + \dots\right)$.

Step 3: Identify and sum the geometric series for A. The odd terms of A form a geometric series: $\frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^5} + \dots$ The first term is $a_{odd,A} = \frac{1}{2}$. The common ratio is $r_{odd,A} = \frac{1/2^3}{1/2} = \frac{1}{4}$. Since $|r_{odd,A}| < 1$, the sum is $\frac{a_{odd,A}}{1-r_{odd,A}} = \frac{1/2}{1-1/4} = \frac{1/2}{3/4} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$.

The even terms of A form a geometric series: $\frac{1}{4^2} + \frac{1}{4^4} + \frac{1}{4^6} + \dots$ The first term is $a_{even,A} = \frac{1}{4^2} = \frac{1}{16}$. The common ratio is $r_{even,A} = \frac{1/4^4}{1/4^2} = \frac{1}{4^2} = \frac{1}{16}$. Since $|r_{even,A}| < 1$, the sum is $\frac{a_{even,A}}{1-r_{even,A}} = \frac{1/16}{1-1/16} = \frac{1/16}{15/16} = \frac{1}{15}$.

Therefore, $A = \frac{2}{3} + \frac{1}{15} = \frac{10}{15} + \frac{1}{15} = \frac{11}{15}$.

Step 4: Identify and sum the geometric series for B. The odd terms of B form a geometric series: $-\frac{1}{2} - \frac{1}{2^3} - \frac{1}{2^5} - \dots$ The first term is $a_{odd,B} = -\frac{1}{2}$. The common ratio is $r_{odd,B} = \frac{-1/2^3}{-1/2} = \frac{1}{4}$. Since $|r_{odd,B}| < 1$, the sum is $\frac{a_{odd,B}}{1-r_{odd,B}} = \frac{-1/2}{1-1/4} = \frac{-1/2}{3/4} = -\frac{1}{2} \times \frac{4}{3} = -\frac{2}{3}$.

The even terms of B form a geometric series: $\frac{1}{4^2} + \frac{1}{4^4} + \frac{1}{4^6} + \dots$ This is the same series as the even terms of A. The first term is $a_{even,B} = \frac{1}{16}$. The common ratio is $r_{even,B} = \frac{1}{16}$. Since $|r_{even,B}| < 1$, the sum is $\frac{a_{even,B}}{1-r_{even,B}} = \frac{1/16}{1-1/16} = \frac{1}{15}$.

Therefore, $B = -\frac{2}{3} + \frac{1}{15} = -\frac{10}{15} + \frac{1}{15} = -\frac{9}{15} = -\frac{3}{5}$.

Step 5: Calculate the ratio A/B. We have $A = \frac{11}{15}$ and $B = -\frac{3}{5}$. The ratio $\frac{A}{B} = \frac{11/15}{-3/5} = \frac{11}{15} \times \frac{5}{-3} = \frac{11}{3} \times \frac{1}{-3} = -\frac{11}{9}$.

Common Mistakes & Tips

• Correctly identifying the first term and common ratio: For geometric series, ensure you are using the correct first term and common ratio, especially when signs change or terms are squared.
• Handling the $(-1)^n$ term: Carefully consider the value of $(-1)^n$ for odd and even $n$ when calculating the terms of the series.
• Arithmetic with fractions: Double-check all fraction addition, subtraction, and multiplication to avoid errors.

Summary

The problem involves evaluating two infinite series, $A$ and $B$. By recognizing that the term $(-1)^n$ causes different behavior for odd and even values of $n$, we can split each series into two separate geometric series. We then apply the formula for the sum of an infinite geometric series to find the values of $A$ and $B$. Finally, we compute the ratio $\frac{A}{B}$. The value of $A$ is $\frac{11}{15}$ and the value of $B$ is $-\frac{3}{5}$, leading to a ratio of $-\frac{11}{9}$.

The final answer is \boxed{-\frac{11}{9}}.